I am confused. Which version of the question is it? The one here, or this one?

If you read carefully, they are the exact same question, just this one is in slightly better English

I knew that. Just trying to get one of them locked up forever! But this one seems to have a problem with its quantifiers. Should it be: Prove for any $M>2$, there exists an increasing sequence of positive integers $a_1<a_2<\ldots $ satisfying: 1) $a_i>M^i$ for any $i$; 2) Given $n\in\mathbb{Z}/\{0\}$, there exists a positive integer $m$ and $b_1,b_2,\ldots ,b_m\in\left\{ -1,1\right\}$, satisfying $n=a_1b_1+a_2b_2+\ldots +a_mb_m$, and there is no such $b_k$ for any $m$ if $n=0$. Because the way it's written now it implies there's an $m$ that works for all $n\neq0$.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=457772&p=2572193#p2572193

Let ${{a}_{1}}={{M}^{2}},{{a}_{2}}={{M}^{2}}+1$, then $1=-{{a}_{1}}+{{a}_{2}}$. For any integer $r\ge 2$, ${{a}_{2r-1}}={{M}^{2r}}+\sum\limits_{i=1}^{2r-2}{{{a}_{i}}}$, and ${{a}_{2r}}=r+\sum\limits_{i=1}^{2r-1}{{{a}_{i}}}$. For any $r\ge 1$, ${{a}_{2r-1}}\ge {{M}^{2r}}>{{M}^{2r-1}}$, and ${{a}_{2r}}>{{a}_{2r-1}}\ge {{M}^{2r}}$. $\pm {{a}_{1}}\ne 0$, for any $m\ge 2$, ${{b}_{1}},{{b}_{2}},...,{{b}_{m}}\in \left\{ -1,1 \right\}$, $\left| {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+...+{{a}_{m}}{{b}_{m}} \right|\ge \left| {{a}_{m}} \right|-\left| {{a}_{1}} \right|-\left| {{a}_{2}} \right|-...\left| {{a}_{m-1}} \right|={{a}_{m}}-\sum\limits_{i=1}^{m-1}{{{a}_{i}}}>0$; for any $n\ge 1$, $\pm n=\pm \left( {{a}_{2n}}-\sum\limits_{i=1}^{2n-1}{{{a}_{i}}} \right)$. So we have done.

yunxiu wrote: Let ${{a}_{1}}={{M}^{2}},{{a}_{2}}={{M}^{2}}+1$, then $1=-{{a}_{1}}+{{a}_{2}}$. For any integer $r\ge 2$, ${{a}_{2r-1}}={{M}^{2r}}+\sum\limits_{i=1}^{2r-2}{{{a}_{i}}}$, and ${{a}_{2r}}=r+\sum\limits_{i=1}^{2r-1}{{{a}_{i}}}$. For any $r\ge 1$, ${{a}_{2r-1}}\ge {{M}^{2r}}>{{M}^{2r-1}}$, and ${{a}_{2r}}>{{a}_{2r-1}}\ge {{M}^{2r}}$. $\pm {{a}_{1}}\ne 0$, for any $m\ge 2$, ${{b}_{1}},{{b}_{2}},...,{{b}_{m}}\in \left\{ -1,1 \right\}$, $\left| {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+...+{{a}_{m}}{{b}_{m}} \right|\ge \left| {{a}_{m}} \right|-\left| {{a}_{1}} \right|-\left| {{a}_{2}} \right|-...\left| {{a}_{m-1}} \right|={{a}_{m}}-\sum\limits_{i=1}^{m-1}{{{a}_{i}}}>0$; for any $n\ge 1$, $\pm n=\pm \left( {{a}_{2n}}-\sum\limits_{i=1}^{2n-1}{{{a}_{i}}} \right)$. So we have done. why $\displaystyle a_{2r-1}\in \mathbb{N}$ ? (we don't know that $\displaystyle M\in \mathbb{N}$)

sorry for my ridicule comment,but for a complete solution you need to use $\left \lceil M \right \rceil$ instead $M$.

For another solution : $a_{2n-1}=M^{2n}$ , $a_{2n}=M^{2n}+2^{n-1}$

Mosquitall wrote: For another solution : $a_{2n-1}=M^{2n}$ , $a_{2n}=M^{2n}+2^{n-1}$ sorry I don't think this is right,can you explain it more detail

Note that $-a$ is obtained from $a$ by changing the signs in the expression $a=a_1b_1+a_2b_2+\ldots +a_kb_k$, so it suffices to prove the problem statement for positive integers. We construct the sequence $(a_n)$ by induction. Let $a_1>M^2$, $a_2=a_1+1$. Suppose we constructed the first $k$ terms. Call an integer attainable if it can be be obtained by the sum $a_1b_1+a_2b_2+\ldots +a_lb_l$, $b_1,b_2,\ldots ,b_l\in\left\{ -1,1\right\}$, $l\le k$, and unattainable in opposite case. Let $m\in\mathbb N$ be the smallest unattainable positive integer. Lemma. There exists an attainable number $a\in\mathbb N$ such that $m+a$ is unattainable. Proof: Suppose not. From the definition of $m$ we get that the numbers $1,2,\ldots, m-1$ are attainable. Then, by our assumption it follows that $m+1,m+2,\ldots, m+(m-1)$ are also attainable. Proceeding by induction, we get that all the numbers not divisible by $m$ are attainable, which is a contradiction since all the numbers greater than $a_1+a_2+\ldots +a_n$ are unattainable.$\blacksquare$ According to Lemma, we can pick an attainable positive integer $a$ such that $m+a$ is unattainable. Then, let $$ a_{k+1}>\max\left\{\sum\limits_{i=1}^k{a_i}, M^{k+2}\right\},\mbox{ } a_{k+2}=a_{k+1}+m+a. $$Then the number $m$ is attainable by the new sequence since $m=-a-a_{k+1}+a_{k+2}$, and $-a$ is attainable ($-a$ is obtained from $a$ by changing the signs in the expression $a=a_1b_1+a_2b_2+\ldots +a_kb_k$). Suppose then that $0$ became attainable after adding $a_{k+1}$ and $a_{k+2}$ into the sequence, i.e. $$ 0=a_1b_1+a_2b_2+\ldots +a_{k+2}b_{k+2},\qquad\mbox{ }b_1,b_2,\ldots ,b_k\in\left\{ -1,1\right\}, $$or $$ a_1c_1+a_2c_2+\ldots +a_{k}c_{k}=a_{k+1}c_{k+1}+a_{k+2}c_{k+2},\qquad\mbox{ }c_1,c_2,\ldots ,c_k\in\left\{ -1,1\right\} $$If $c_{k+1}=c_{k+2}\ne0$ or $c_{k+1}\ne0$, $c_{k+2}\ne0$, then $|a_{k+1}c_{k+1}+a_{k+2}c_{k+2}| > a_1+a_2+\ldots+a_k\ge |a_1c_1+a_2c_2+\ldots +a_{k}c_{k}|$, a contradiction. Then $c_{k+1}=-c_{k+2}\ne0$, and $|a_{k+1}c_{k+1}+a_{k+2}c_{k+2}|=m+a=|a_1c_1+a_2c_2+\ldots +a_{k}c_{k}|$, which is a contradiction since $m+a$ is unattainable. If $c_{k+1}=c_{k+2}=0$, then again contradiction since $0\ne a_1c_1+a_2c_2+\ldots +a_{k}c_{k}$. This shows that our induction step is correct, and hence the sequence $(a_n)_{n\in\mathbb N}$ will satisfy the problem statement.

If it is, in my opinion this is too easy for a CMO3, which is quite surprising given the difficulty of day 2...

Pick $a_1=M+1$. Suppose we have define $a_1,...,a_m$, Let $$T_m=\{n:n=b_1a_1+...+b_ma_m:b_1,...,b_m\in\{-1,1\}\}$$Let $$S_m=T_1\cup...\cup T_m$$Let $k$ be the integer with the smallest absolute value that is not in $S_m$. Pick $a_{m+1}$ to be sufficiently large, let $\lceil M\rceil +1=k$ and let $r$ be an integer such that $k^r>M^{r-1}$, then define $$a_{m+i}=(\lceil M\rceil+1)a_{m+i-1}$$for each $2\leq i\leq r+1$. Now we define $$a_{m+r+1}=a_1+...+a_{m+r}+|k|$$By construction we see that $$-(a_1+...+a_{m+r}+a_{m+r+1}=|k|$$$$a_{m+r+1}>M^{m+r+1}$$and $$0\notin S_{m+r+1}$$as desired.